[proofplan] The Hurwitz hypothesis gives exponential decay of the semigroup $e^{tA}$, so the matrix-valued energy density $e^{tA^\top}C^\top C e^{tA}$ is integrable on $[0,\infty)$. Integrating the derivative of $e^{tA^\top}C^\top C e^{tA}$ over finite horizons yields the Lyapunov equation after the boundary term at infinity vanishes. Uniqueness follows by applying the same decay argument to the homogeneous Lyapunov equation. Finally, the quadratic form of $W_o$ is exactly the total output energy $\int_0^\infty |Ce^{tA}x|^2\,d\mathcal{L}^1(t)$, so positive definiteness is equivalent to the absence of nonzero states with identically zero output, which is precisely observability by the [Cayley-Hamilton theorem](/theorems/865). [/proofplan]

[step:Use Hurwitz decay to prove the Gramian integral exists] For a matrix $B \in \mathbb{R}^{n \times n}$, let $\|B\|_{\mathrm{op}} := \sup\{|Bx| : x \in \mathbb{R}^n, |x| = 1\}$ denote the Euclidean operator norm. Since $A$ is Hurwitz, the finite-dimensional semigroup estimate obtained by applying the [Jordan normal form](/theorems/864) to the complexification of $A$ gives constants $M>0$ and $\alpha>0$ such that \begin{align*} \|e^{tA}\|_{\mathrm{op}} \leq M e^{-\alpha t} \end{align*} for every $t \geq 0$. Indeed, over $\mathbb{C}$ each Jordan block contributes a factor bounded by $q(t)e^{-\beta t}$, where $q: [0,\infty) \to [0,\infty)$ is a polynomial and $\beta>0$ is smaller than the absolute value of the real part of the corresponding eigenvalue. Choosing $0<\alpha<\beta$ gives \begin{align*} q(t)e^{-\beta t}=q(t)e^{-(\beta-\alpha)t}e^{-\alpha t}\leq M_q e^{-\alpha t} \end{align*} for all $t\geq 0$, where $M_q:=\sup_{s\geq 0} q(s)e^{-(\beta-\alpha)s}<\infty$. Taking the maximum of the finitely many block constants gives the stated estimate for $e^{tA}$. Define the matrix-valued function $G: [0,\infty) \to \mathbb{R}^{n \times n}$ by \begin{align*} G(t) := e^{tA^\top}C^\top C e^{tA} \quad \text{for } t \geq 0. \end{align*} The function $G$ is continuous because the matrix exponential is continuous. For every $t \geq 0$, \begin{align*} \|G(t)\|_{\mathrm{op}} \leq \|e^{tA^\top}\|_{\mathrm{op}} \|C^\top C\|_{\mathrm{op}} \|e^{tA}\|_{\mathrm{op}}. \end{align*} Since $\|e^{tA^\top}\|_{\mathrm{op}}=\|e^{tA}\|_{\mathrm{op}}$, this gives \begin{align*} \|G(t)\|_{\mathrm{op}} \leq M^2 \|C^\top C\|_{\mathrm{op}} e^{-2\alpha t}. \end{align*} The scalar function $t \mapsto M^2\|C^\top C\|_{\mathrm{op}}e^{-2\alpha t}$ is integrable on $[0,\infty)$ with respect to $\mathcal{L}^1$. Therefore each entry of $G$ is absolutely integrable, and \begin{align*} W_o := \int_0^\infty G(t)\, d\mathcal{L}^1(t) \end{align*} exists entrywise. [/step]

[step:Show that the Gramian is symmetric and positive semidefinite]For every $t \geq 0$, \begin{align*} G(t)^\top = \left(e^{tA^\top}C^\top C e^{tA}\right)^\top = e^{tA^\top}C^\top C e^{tA}=G(t), \end{align*} so integration preserves symmetry and $W_o^\top=W_o$. Let $x \in \mathbb{R}^n$. Define the map $y_x: [0,\infty) \to \mathbb{R}^m$ by \begin{align*} y_x(t) := C e^{tA}x \quad \text{for } t \geq 0. \end{align*} The map $y_x$ is continuous. Using the definition of $W_o$ and associativity of matrix multiplication, \begin{align*} x^\top W_o x = \int_0^\infty x^\top e^{tA^\top}C^\top C e^{tA}x\, d\mathcal{L}^1(t). \end{align*} The integrand is the Euclidean square norm of $y_x(t)$: \begin{align*} x^\top e^{tA^\top}C^\top C e^{tA}x = |Ce^{tA}x|^2. \end{align*} Hence \begin{align*} x^\top W_o x = \int_0^\infty |Ce^{tA}x|^2\, d\mathcal{L}^1(t) \geq 0. \end{align*} Thus $W_o$ is positive semidefinite.[/step]

[guided]The matrix $G(t)=e^{tA^\top}C^\top C e^{tA}$ has the form $B(t)^\top B(t)$ with $B(t)=Ce^{tA}$. This is the structural reason the Gramian is positive semidefinite: it measures output energy. We also record the integrability needed for the entrywise improper integral. The Hurwitz decay estimate gives constants $M>0$ and $\alpha>0$ such that $\|e^{tA}\|_{\mathrm{op}}\leq Me^{-\alpha t}$ for all $t\geq 0$. Hence \begin{align*} \|G(t)\|_{\mathrm{op}}\leq \|e^{tA^\top}\|_{\mathrm{op}}\|C^\top C\|_{\mathrm{op}}\|e^{tA}\|_{\mathrm{op}}\leq M^2\|C^\top C\|_{\mathrm{op}}e^{-2\alpha t}. \end{align*} The dominating scalar function $t\mapsto M^2\|C^\top C\|_{\mathrm{op}}e^{-2\alpha t}$ is integrable on $[0,\infty)$ with respect to $\mathcal{L}^1$, so every entry of $G$ is absolutely integrable and the entrywise improper integral defining $W_o$ exists. First, we check symmetry. For each $t \geq 0$, \begin{align*} G(t)^\top = \left(e^{tA^\top}C^\top C e^{tA}\right)^\top. \end{align*} Using $(PQ)^\top=Q^\top P^\top$ and $(e^{tA})^\top=e^{tA^\top}$, we obtain \begin{align*} G(t)^\top = e^{tA^\top}C^\top C e^{tA}=G(t). \end{align*} Since every integrand matrix is symmetric and the integral is taken entrywise, the integral matrix $W_o$ is symmetric. Now let $x \in \mathbb{R}^n$. Define the output trajectory generated by the initial state $x$ as the continuous map $y_x: [0,\infty) \to \mathbb{R}^m$ given by \begin{align*} y_x(t) := C e^{tA}x \quad \text{for } t \geq 0. \end{align*} Then the quadratic form of $W_o$ at $x$ is \begin{align*} x^\top W_o x = \int_0^\infty x^\top e^{tA^\top}C^\top C e^{tA}x\, d\mathcal{L}^1(t). \end{align*} The scalar inside the integral can be rewritten as a Euclidean norm: \begin{align*} x^\top e^{tA^\top}C^\top C e^{tA}x = (Ce^{tA}x)^\top(Ce^{tA}x)=|Ce^{tA}x|^2. \end{align*} Therefore \begin{align*} x^\top W_o x = \int_0^\infty |Ce^{tA}x|^2\, d\mathcal{L}^1(t). \end{align*} The integrand is nonnegative for every $t \geq 0$, so the integral is nonnegative. Since this holds for every $x \in \mathbb{R}^n$, the matrix $W_o$ is positive semidefinite.[/guided]

[step:Integrate the derivative to obtain the Lyapunov equation] For $T>0$, define the finite-horizon matrix \begin{align*} W_T := \int_0^{\!T}e^{tA^\top}C^\top C e^{tA}\, d\mathcal{L}^1(t). \end{align*} The product rule and the identities $\frac{d}{dt}e^{tA}=Ae^{tA}=e^{tA}A$ and $\frac{d}{dt}e^{tA^\top}=A^\top e^{tA^\top}=e^{tA^\top}A^\top$ give \begin{align*} \frac{d}{dt}\left(e^{tA^\top}C^\top C e^{tA}\right) = A^\top e^{tA^\top}C^\top C e^{tA} + e^{tA^\top}C^\top C e^{tA}A. \end{align*} Integrating this identity over $[0,T]$ with respect to $\mathcal{L}^1$ gives \begin{align*} A^\top W_T + W_T A = e^{TA^\top}C^\top C e^{TA}-C^\top C. \end{align*} As $T \to \infty$, the exponential decay estimate gives \begin{align*} \|e^{TA^\top}C^\top C e^{TA}\|_{\mathrm{op}} \leq M^2\|C^\top C\|_{\mathrm{op}}e^{-2\alpha T} \to 0. \end{align*} Also $W_T \to W_o$ entrywise by the definition of the improper integral. Since left and right multiplication by the fixed matrices $A^\top$ and $A$ are entrywise continuous linear maps on the finite-dimensional [vector space](/page/Vector%20Space) $\mathbb{R}^{n \times n}$, we have $A^\top W_T + W_TA \to A^\top W_o + W_oA$ entrywise. Passing to the limit yields \begin{align*} A^\top W_o + W_o A = -C^\top C. \end{align*} [/step]

[step:Prove uniqueness by solving the homogeneous Lyapunov equation] Let $W_1,W_2 \in \mathbb{R}^{n \times n}$ both satisfy \begin{align*} A^\top W_i + W_i A = -C^\top C \end{align*} for $i \in \{1,2\}$. Define $H:=W_1-W_2$. Then \begin{align*} A^\top H + H A = 0. \end{align*} Define the matrix-valued function $F_H: [0,\infty) \to \mathbb{R}^{n \times n}$ by \begin{align*} F_H(t) := e^{tA^\top}H e^{tA} \quad \text{for } t \geq 0. \end{align*} By the product rule, \begin{align*} F_H'(t)=e^{tA^\top}(A^\top H+HA)e^{tA}=0. \end{align*} Thus $F_H(t)=F_H(0)=H$ for every $t \geq 0$. On the other hand, \begin{align*} \|F_H(t)\|_{\mathrm{op}} \leq M^2\|H\|_{\mathrm{op}}e^{-2\alpha t} \to 0. \end{align*} Therefore $H=0$, so $W_1=W_2$. Hence $W_o$ is the unique solution of the Lyapunov equation. [/step]

[step:Identify positive definiteness with observability] Let $\mathcal{O}: \mathbb{R}^n \to \mathbb{R}^{mn}$ be the observability map defined by \begin{align*} \mathcal{O}(x) := (Cx, CAx, \dots, CA^{n-1}x) \quad \text{for } x \in \mathbb{R}^n, \end{align*} where the tuple is identified with an element of $\mathbb{R}^{mn}$ by stacking its $n$ components in order. By the finite-dimensional definition of observability, the pair $(C,A)$ is observable exactly when $\ker \mathcal{O}=\{0\}$. Assume first that $(C,A)$ is not observable. Then there exists $x \in \mathbb{R}^n$ with $x \neq 0$ and \begin{align*} CA^k x=0 \end{align*} for every $k \in \{0,\dots,n-1\}$. By the [Cayley-Hamilton Theorem](/theorems/586), every power $A^k$ with $k \geq n$ is a linear combination of $I,A,\dots,A^{n-1}$, so $CA^k x=0$ for every $k \geq 0$. Expanding the matrix exponential in its absolutely convergent [power series](/page/Power%20Series) gives \begin{align*} Ce^{tA}x = \sum_{k=0}^\infty \frac{t^k}{k!}CA^k x = 0 \end{align*} for every $t \geq 0$. Therefore \begin{align*} x^\top W_o x = \int_0^\infty |Ce^{tA}x|^2\, d\mathcal{L}^1(t)=0. \end{align*} Since $x \neq 0$, $W_o$ is not positive definite. Conversely, assume $W_o$ is not positive definite. Since $W_o$ is positive semidefinite, there exists $x \in \mathbb{R}^n$ with $x \neq 0$ and $x^\top W_o x=0$. Hence \begin{align*} \int_0^\infty |Ce^{tA}x|^2\, d\mathcal{L}^1(t)=0. \end{align*} The function $t \mapsto |Ce^{tA}x|^2$ is continuous and nonnegative, so it must vanish for every $t \geq 0$: if it were positive at some $t_0 \geq 0$, continuity would give a neighbourhood of $t_0$ on which it is bounded below by a positive constant, and that neighbourhood has positive $\mathcal{L}^1$-measure in $[0,\infty)$. Thus $Ce^{tA}x=0$ for every $t \geq 0$. Differentiating the map $t \mapsto Ce^{tA}x$ at $t=0$ repeatedly gives \begin{align*} CA^k x=0 \end{align*} for every $k \in \{0,\dots,n-1\}$. Hence $x \in \ker \mathcal{O}$ and $x \neq 0$, so $(C,A)$ is not observable. We have proved that $(C,A)$ fails to be observable exactly when $W_o$ fails to be positive definite. Equivalently, $(C,A)$ is observable if and only if $W_o$ is positive definite. [/step]